Show that $F(G(x))=G(F(x)).$ 
Let $V$ be a finitely dimensional linear space, and $F:V\rightarrow V$
  and $G:V\rightarrow V$ linear transformations that are diagonalizable,
  that is: there exists a base $\mathbf{e}$ for $V$ such that the matrix
  to $F$ in the basis $\mathbf{e}$ is diagonal and a base $\mathbf{f}$
  for $V$ such that the matrix to $G$ is diagonal.
i) Show that if $F$ and $G$ are simultaneously diagonalizable, that is if $\mathbf{e}=\mathbf{f},$ then it follows that
$$F(G(x))=G(F(x)), \quad \forall x\in V. \tag{1}$$
ii) Show that if $(1)$ holds, then there exists a basis $\mathbf{g}$ (possibly different from both $\mathbf{e}$ and $\mathbf{f}$) such that
  the matrices to both $F$ and $G$ in the basis $\mathbf{g}$ are
  diagonal.

Okay, I usually post good questions where I've done my own work before asking for help. But this one has thrown me away. I seriously don't even know how to begin or what I should proove.
This is no homework or assignment, I'm just doing practice problems for my exam at the end of this month. Any tips/tricks/translation of the problem statement is welcome!
 A: If $A$ and $B$ are matrices for $F$ and $G$ with respect to base $e=f$, then there exists an invertible matrix $P$ such $A=P\Delta_AP^{-1}$ and $B=P\Delta_BP^{-1}$, where $\Delta_A,\Delta_B$ are diagonal matrices. Then $$AB=P\Delta_A\Delta_BP^{-1}\\BA=P\Delta_B\Delta_AP^{-1}$$
Now from the fact that, any two diagonal matrices commute we get $A$ and $B$ also commute.
A: Regarding part 1: If $F$ and $G$ are simultaneously diagonalizable, then with respect to some basis of $V$ we can represent $F$ and $G$ as
\begin{align*}
F = \left( \hspace{-0.4cm}
\begin{matrix} 
&\lambda_1 & &0 & &\cdots & &0\\
&0 & &\lambda_2 & &\cdots & &0\\
&\vdots & &\vdots & &\ddots & &\vdots\\
&0 & &0 & &\cdots & &\lambda_n\\
\end{matrix} \right), \quad G = \left( \hspace{-0.4cm}
\begin{matrix} 
&\kappa_1 & &0 & &\cdots & &0\\
&0 & &\kappa_2 & &\cdots & &0\\
&\vdots & &\vdots & &\ddots & &\vdots\\
&0 & &0 & &\cdots & &\kappa_n\\
\end{matrix} \right),
\end{align*}
the point being that both transformations can be represented as diagonal matrices with respect to the same basis. But then, since $F$ and $G$ are diagonal matrices, they commute:
\begin{align*}
FG &= \left( \hspace{-0.4cm}
\begin{matrix} 
&\lambda_1 & &0 & &\cdots & &0\\
&0 & &\lambda_2 & &\cdots & &0\\
&\vdots & &\vdots & &\ddots & &\vdots\\
&0 & &0 & &\cdots & &\lambda_n\\
\end{matrix} \right) \left( \hspace{-0.4cm}
\begin{matrix} 
&\kappa_1 & &0 & &\cdots & &0\\
&0 & &\kappa_2 & &\cdots & &0\\
&\vdots & &\vdots & &\ddots & &\vdots\\
&0 & &0 & &\cdots & &\kappa_n\\
\end{matrix} \right)\\
&= \left( \hspace{-0.4cm}
\begin{matrix} 
&\lambda_1 \kappa_1 & &0 & &\cdots & &0\\
&0 & &\lambda_2 \kappa_2 & &\cdots & &0\\
&\vdots & &\vdots & &\ddots & &\vdots\\
&0 & &0 & &\cdots & &\lambda_n \kappa_n\\
\end{matrix} \right) \\
&= \left( \hspace{-0.4cm}
\begin{matrix} 
&\kappa_1 & &0 & &\cdots & &0\\
&0 & &\kappa_2 & &\cdots & &0\\
&\vdots & &\vdots & &\ddots & &\vdots\\
&0 & &0 & &\cdots & &\kappa_n\\
\end{matrix} \right) \left( \hspace{-0.4cm}
\begin{matrix} 
&\lambda_1 & &0 & &\cdots & &0\\
&0 & &\lambda_2 & &\cdots & &0\\
&\vdots & &\vdots & &\ddots & &\vdots\\
&0 & &0 & &\cdots & &\lambda_n\\
\end{matrix} \right) = GF.
\end{align*}
A: For 1, if $e\in\mathbf e$, we have $Fe=\alpha_e e$, $Ge=\beta_e e$ for some coefficients $\alpha_e,\beta_e$. Then 
$$
FGe=\beta_e Fe=\alpha_e\beta_e e=\alpha_e Ge=GFe. 
$$
So $FG$ and $GF$ agree on all elements of a basis, and thus are equal. 
For 2, write $\mathbf e=\{e_1,\ldots,e_n\}$. By hypothesis, $Fe_j=\alpha_j e_j$, $j=1,\ldots,n$. We have 
$$
FGe_j=GFe_j=\alpha_jGe_j.
$$
So $Ge_j$ is an eigenvector for $F$ with eigenvalue $\alpha_j$. That is, the eigenspaces of $F$ are invariant for $G$. That is, if $E_1,\ldots,E_r$ are the eigenspaces of $F$ corresponding to distinct eigenvalues, $GE_j\subset E_j$. So we may consider $G$ as a linear transformation on $E_j$. 
Now we want to show that $G$ is diagonalizable on $E_j$. The key fact is that "diagonalizable" is equivalent to the fact that the minimal polynomial has no repeated roots. So, since $G$ is diagonalizable, $q(G)=0$, where  $q(t)=(t-\beta_1)\cdots(t-\beta_n)$. Now let $p_j$ be the minimal polynomial of $G|_{E_j}$. Since $q(G)=0$, we must have $p_j|q$, and so the roots of $p_j$ are simple. Thus, $G|_{E_j}$ is diagonalizable. 
So, for each $E_j$ we have a basis $e_{1j},\ldots,e_{m_jj}$ of eigenvectors for $G$. As $e_{kj}\in E_j$, it is also an eigenvector for $F$. So if we let 
$$
\mathbf g=\{e_{11},\ldots,e_{m_11},e_{21},\ldots,e_{m_22},\ldots,e_{r1},\ldots,e_{m_rr}\},
$$
we have a basis where its elements are both eigenvectors for $F$ and for $G$. 
